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At $15^{\circ} \mathrm{C}$ and $900 \mathrm{~mm}$ pressure, $0.4$ gram of a certain gas occupied $3280 \mathrm{~mL}$. If the pressure is changed to $1300\mathrm{~mm}$ at constant temperature. What is the density of the gas?
Solution:
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Verified Answer
Density $=\frac{\text { Mass }}{\text { Volume }}$
$\therefore \quad$ Density at $900 \mathrm{~mm}$ pressure
$\begin{aligned}
&=\frac{0.4}{3280} \mathrm{~g} / \mathrm{mL} \\
&=0.000122 \mathrm{~g} / \mathrm{mL}
\end{aligned}$
At constant temperature, Density $\times$ Pressure
$\therefore \quad \frac{d_1}{P_1}=\frac{d_2}{P_2} \quad \text { or } \quad d_2=\frac{d_1 \times d_2}{P_1}$
$\begin{array}{ll}\mathrm{d}_1=0.000122 \mathrm{~g} / \mathrm{mL} & \mathrm{d}_2=? \\ \mathrm{P}_1=900 \mathrm{~mm} & \mathrm{P}_2=1300 \mathrm{~mm}\end{array}$
$\therefore \quad \mathrm{d}_2=\frac{0.000122 \times 1300}{900}=0.0001762 \mathrm{~g} / \mathrm{mL}$
Hence, density of gas at $1300 \mathrm{~mm}=0.0001762 \mathrm{~g} / \mathrm{mL}$.
$\therefore \quad$ Density at $900 \mathrm{~mm}$ pressure
$\begin{aligned}
&=\frac{0.4}{3280} \mathrm{~g} / \mathrm{mL} \\
&=0.000122 \mathrm{~g} / \mathrm{mL}
\end{aligned}$
At constant temperature, Density $\times$ Pressure
$\therefore \quad \frac{d_1}{P_1}=\frac{d_2}{P_2} \quad \text { or } \quad d_2=\frac{d_1 \times d_2}{P_1}$
$\begin{array}{ll}\mathrm{d}_1=0.000122 \mathrm{~g} / \mathrm{mL} & \mathrm{d}_2=? \\ \mathrm{P}_1=900 \mathrm{~mm} & \mathrm{P}_2=1300 \mathrm{~mm}\end{array}$
$\therefore \quad \mathrm{d}_2=\frac{0.000122 \times 1300}{900}=0.0001762 \mathrm{~g} / \mathrm{mL}$
Hence, density of gas at $1300 \mathrm{~mm}=0.0001762 \mathrm{~g} / \mathrm{mL}$.
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